In this video, I work through an example of proving that a relation is an equivalence relation. We do this by showing that the relation is reflexive, symmetric, and transitive.
Sorry i cant understand the part where a-b = 3k, where a-b is divisible by 3. May I know how did you get 3k? as I thought it will be k/3 instead. Sorry!
in case you didn't find out yet, that's a division with elements replaced. like if (15/k) = 3 -> 15 = 3k -> k = 5. so in this problem x - y = 3k, you can see as (x-y)/3 = k. if k is in Z, x-y is divisible by 3.
Hey may I ask a question: let’s say we have an equivalence relation aRb. Why can’t I represent this within set theory as set T comprising subset of Cartesian product of a and b, mapped to a set U which contains true or false? Thanks so much!!
@@HamblinMath hey friend! Sorry for not understanding but would you unpack your reply a bit? I don’t understand why people on Reddit told me relations like equivalence or just symmetrical or just reflexive are “meta” relations and can’t really be seen as relations between two sets and set theory doesn’t allow it.
@@HamblinMath to clarify my second reply to your reply: but I would very much like to know how we can do this with the truth/false as elements of the destination set!
@@MathCuriousity You *can* define a relation as a function from A x A to {True, False}, but I don't see any reason why you would, since the relation itself would be just the preimage of "True." The function doesn't gain you anything.
@@HamblinMath I don’t understand - the whole confusion I have is -if I have a reflexive relation for instance - it seems the ordered pair is of the elements a and b the relation acts on - but where is the “truth” stored ?
